Nothing cool about the last digits

Let $A = 9^{9^9} \\ \implies \log_{10} A = 9^9 \log_{10} 9$

By computation,

1
2
3

In[12]:= N[9^9 Log10[9], 32]

Out[12]= 3.6969309963157035874354309509548*10^8

We have,

$\log_{10} A = 9^9 \log_{10} 9 = 3.6969309963157035874354309509548 \times 10^8$

Raising both sides to the power of 10,

$\implies A = 10 ^ { 3.6969309963157035874354309509548 \times 10^8 }$

By computation,

1
2
3

In[13]:= N[10^(3.6969309963157035874354309509548 10^8), 32]

Out[13]= 4.2812477317574704803698*10^369693099

So, $A = 4.2812477317574704803698 \times 10^{369693099}$

(Similar idea to Agnishom Chattopadhyay)

Let $N=9^{9^9}=\overline{a_1 \cdots a_n b_1 \cdots b_m}$ . Observe that $\overline{a_1 \cdots a_n }\cdot 10^m\leq N < \overline{a_1 \cdots a_n}\cdot 10^{m+1}$ Then we get $\log_{10} \overline{a_1 \cdots a_n }+m\leq \log_{10} N<\log_{10} \overline{a_1 \cdots a_n }+m+1$ which implies $(\text{fractional part of }\log_{10} N)=\log_{10} \overline{a_1 \cdots a_n } -(n-1)$
This makes the calculation faster.

• Code example (in python):

1
2
3
4
5
6

from math import log10 as log

def first_n_digits(log_N, n): # log_N = value of log(N)
    return int(10**(log_N-int(log_N)+n-1))

print first_n_digits((9**9)*log(9), 3)

• Output :

1

428

Press [ 9 ][ * ][ = ][ = ][ = ]... or [ 9 ][ * ][ * ][ = ][ = ][ = ].. depends onto type of calculator we are using, observe changes of first 3 digits.

There is valuable theory being identified within this week. To tell a hint, first few digits belong to propagating figures while last few digits belong to no squeezed out requirement. The more the better and the least the better respectively but the demand may go against them.

1
2
3
4
5

428124773175749798 [18]
42812477029460678 [17]
427836765123 [12]
218540170 [9]
54359 [5]

We can realized from analysis that 428124... could be the first 6 digits. A range from 9 to 18 is an option of length required for using extract-able 18 S.F. real numbers.

$9^{9^9} = 10^{387420489 \log 9}$ = $10^{369693099.63157035874354309509548}$ = $10^{369693099} \times 10^{0.63157035874354309509548}$

= $4.2812477317574704803698711592765 \times 10^{369693099}$

Scientific method which is useful is proven by mathematical method.

Answer: $\boxed{428}$